To solve this problem, we need to determine the distance traveled by a freely falling body during each second of its fall.$\\$The distance $s_n$ traveled by a freely falling body in the $n^{th}$ second is given by:$\\ s_n = u + \frac{1}{2} g (2n - 1) \\$For a freely falling body, the initial velocity $u = 0.$ Therefore, the equation simplifies to:$\\ s_n = \frac{1}{2} g (2n - 1) \\$Let's calculate the distance for each second:$\\$• For the 1st second $(n = 1): \\ s_1 = \frac{1}{2} g (2 \cdot 1 - 1) = \frac{1}{2} g \cdot 1 = \frac{1}{2} g \\$• For the 2nd second $(n = 2): \\ s_2 = \frac{1}{2} g (2 \cdot 2 - 1) = \frac{1}{2} g \cdot 3 = \frac{3}{2} g \\$• For the 3rd second $(n = 3): \\ s_3 = \frac{1}{2} g (2 \cdot 3 - 1) = \frac{1}{2} g \cdot 5 = \frac{5}{2} g \\$• For the 4th second $(n = 4): \\ s_4 = \frac{1}{2} g (2 \cdot 4 - 1) = \frac{1}{2} g \cdot 7 = \frac{7}{2} g \\$Now, let's find the ratio of these distances:$\\$Ratio $= s_1 : s_2 : s_3 : s_4 = \frac{1}{2} g : \frac{3}{2} g : \frac{5}{2} g : \frac{7}{2} g \\$Simplifying the ratio by dividing each term by $\frac{1}{2} g: \\$Ratio $= 1 : 3 : 5 : 7 \\$Therefore, the correct option is $3.$